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Kalor
diketahui :
α = 0.000012 /°C
T₁ = 30°C
Lo = 50 cm
L = 50.011 cm
Ditanya :
Besar kenaikan suhu?
Jawab
step 1 : menentukan suhu akhir
L = Lo (1 + α ΔT)
50.011 cm = 50 cm (1 + (0.000012 /°C . (T₂ - 30°C)))
50.011 cm ÷ 50 cm = 1 + (0.000012 /°C . (T₂ - 30°C))
1.00022 = 1 + (0.000012 /°C . (T₂ - 30°C))
1.00022 - 1 = 0.000012 /°C . (T₂ - 30°C)
0.00022 = 0.000012 /°C . (T₂ - 30°C)
0.00022 ÷ 0.000012 /°C = T₂ - 30°C
18°C = T₂ - 30°C
T₂ = 18°C + 30°C
T₂ = 48°C
step 2 : menentukan besar kenaikan suhu
ΔT = T₂ - T₁
ΔT = 48°C - 30°C
ΔT = 18°C
50.011 cm - 50 cm =
0.011 cm = 0,0006.x
(x-30) =
x =